Given two integer arrays, add their elements into third array by satisfying following constraints –
1. Addition should be done starting from 0th index of both arrays.
2. Split the sum if it is a not a single digit number and store the digits in adjacent locations in output array.
3. Output array should accommodate any remaining digits of larger input array.
Examples:
Input: a = [9, 2, 3, 7, 9, 6] b = [3, 1, 4, 7, 8, 7, 6, 9] Output: [1, 2, 3, 7, 1, 4, 1, 7, 1, 3, 6, 9] Input: a = [9343, 2, 3, 7, 9, 6] b = [34, 11, 4, 7, 8, 7, 6, 99] Output: [9, 3, 7, 7, 1, 3, 7, 1, 4, 1, 7, 1, 3, 6, 9, 9] Input: a = [] b = [11, 2, 3 ] Output: [1, 1, 2, 3 ] Input: a = [9, 8, 7, 6, 5, 4, 3, 2, 1] b = [1, 2, 3, 4, 5, 6, 7, 8, 9] Output: [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0]
Difficulty Level: Rookie
The idea is very simple. We maintain an output array and run a loop from the 0th index of both arrays. For each iteration of loop, we consider next elements in both arrays and add them. If the sum is greater than 9, we push the individual digits of the sum to output array else we push the sum itself. Finally we push the remaining elements of larger input array to output array.
Below is the implementation of above idea –
C++
// C++ program to add two arrays following given // constraints #include<bits/stdc++.h> using namespace std; // Function to push individual digits of a number // to output vector from left to right void split(int num, vector<int> &out) { vector<int> arr; while (num) { arr.push_back(num%10); num = num/10; } // reverse the vector arr and append it to output vector out.insert(out.end(), arr.rbegin(), arr.rend()); } // Function to add two arrays keeping given // constraints void addArrays(int arr1[], int arr2[], int m, int n) { // create a vector to store output vector<int> out; // maintain a variable to store current index in // both arrays int i = 0; // loop till arr1 or arr2 runs out while (i < m && i < n) { // read next elements from both arrays and // add them int sum = arr1[i] + arr2[i]; // if sum is single digit number if (sum < 10) out.push_back(sum); else { // if sum is not a single digit number, push // individual digits to output vector split(sum, out); } // increment to next index i++; } // push remaining elements of first input array // (if any) to output vector while (i < m) split(arr1[i++], out); // push remaining elements of second input array // (if any) to output vector while (i < n) split(arr2[i++], out); // print the output vector for (int x : out) cout << x << " "; } // Driver code int main() { int arr1[] = {9343, 2, 3, 7, 9, 6}; int arr2[] = {34, 11, 4, 7, 8, 7, 6, 99}; int m = sizeof(arr1) / sizeof(arr1[0]); int n = sizeof(arr2) / sizeof(arr2[0]); addArrays(arr1, arr2, m, n); return 0; } |
Java
// Java program to add two arrays following given // constraints import java.util.Vector; class GFG { // Function to push individual digits of a number // to output vector from left to right static void split(int num, Vector<Integer> out) { Vector<Integer> arr = new Vector<>(); while (num > 0) { arr.add(num % 10); num /= 10; } // reverse the vector arr and // append it to output vector for (int i = arr.size() - 1; i >= 0; i--) out.add(arr.elementAt(i)); } // Function to add two arrays keeping given // constraints static void addArrays(int[] arr1, int[] arr2, int m, int n) { // create a vector to store output Vector<Integer> out = new Vector<>(); // maintain a variable to store // current index in both arrays int i = 0; // loop till arr1 or arr2 runs out while (i < m && i < n) { // read next elements from both arrays // and add them int sum = arr1[i] + arr2[i]; // if sum is single digit number if (sum < 10) out.add(sum); else // if sum is not a single digit number, // push individual digits to output vector split(sum, out); // increment to next index i++; } // push remaining elements of first input array // (if any) to output vector while (i < m) split(arr1[i++], out); // push remaining elements of second input array // (if any) to output vector while (i < n) split(arr2[i++], out); // print the output vector for (int x : out) System.out.print(x + " "); } // Driver Code public static void main(String[] args) { int[] arr1 = { 9343, 2, 3, 7, 9, 6 }; int[] arr2 = { 34, 11, 4, 7, 8, 7, 6, 99 }; int m = arr1.length; int n = arr2.length; addArrays(arr1, arr2, m, n); } } // This code is contributed by // sanjeev2552 |
C#
// C# program to add two arrays following given // constraints using System; using System.Collections.Generic; class GFG { // Function to push individual digits of a number // to output vector from left to right static void split(int num, List<int> outs) { List<int> arr = new List<int>(); while (num > 0) { arr.Add(num % 10); num /= 10; } // reverse the vector arr and // append it to output vector for (int i = arr.Count - 1; i >= 0; i--) outs.Add(arr[i]); } // Function to add two arrays keeping given // constraints static void addArrays(int[] arr1, int[] arr2, int m, int n) { // create a vector to store output List<int> outs = new List<int>(); // maintain a variable to store // current index in both arrays int i = 0; // loop till arr1 or arr2 runs out while (i < m && i < n) { // read next elements from both arrays // and add them int sum = arr1[i] + arr2[i]; // if sum is single digit number if (sum < 10) outs.Add(sum); else // if sum is not a single digit number, // push individual digits to output vector split(sum, outs); // increment to next index i++; } // push remaining elements of first input array // (if any) to output vector while (i < m) split(arr1[i++], outs); // push remaining elements of second input array // (if any) to output vector while (i < n) split(arr2[i++], outs); // print the output vector foreach (int x in outs) Console.Write(x + " "); } // Driver Code public static void Main(String[] args) { int[] arr1 = { 9343, 2, 3, 7, 9, 6 }; int[] arr2 = { 34, 11, 4, 7, 8, 7, 6, 99 }; int m = arr1.Length; int n = arr2.Length; addArrays(arr1, arr2, m, n); } } // This code is contributed by PrinciRaj1992 |
Output:
9 3 7 7 1 3 7 1 4 1 7 1 3 6 9 9
Time complexity of above solution is O(m + n) as we traverses both arrays exactly once.
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